How do you solve #2x^2+5x-12>=0#?

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Answer 1 · 2016-11-30T15:05:14

The answer is #x in ] -oo,-4] uu [3/2, oo[#

Let's find the roots of the equation

#2x^2+5x-12=0#

This is a simultaneous equation, #ax^2+bx+c=0#

We calculate the discriminant,

#Delta=b^2-4ac=25-4*2*-12=121#

#delta>0#, so we have 2 real roots

The roots are

#x=(-b+-sqrtDelta)/(2a)#

#=(-5+-11)/4#

#x_1=-4# and #x_2=6/4=3/2#

Let #f(x)=2x^2+5x-12#

We do a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-4##color(white)(aaaa)##3/2##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+4##color(white)(aaaaaa)##-##color(white)(aaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-3/2##color(white)(aaaaa)##-##color(white)(aaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##+##color(white)(aaa)##-##color(white)(aaaa)##+#

We need #f(x)>=0#

#x in ] -oo,-4] uu [3/2, oo[#
graph{2x^2+5x-12 [-12.66, 12.66, -6.34, 6.32]}